Question 9.SP.6: Determine the product of inertia of the right triangle shown...
Determine the product of inertia of the right triangle shown (a) with respect to the x and y axes and (b) with respect to centroidal axes parallel to the x and y axes.
STRATEGY: You can approach this problem by using a vertical differential strip element. Because each point of the strip is at a different distance from the x axis, it is necessary to describe this strip mathemati-cally using the parallel-axis theorem. Once you have completed the solu-tion for the product of inertia with respect to the x and y axes, a second application of the parallel-axis theorem yields the product of inertia with respect to the centroidal axes.
MODELING and ANALYSIS:
a. Product of Inertia I_{xy}. Choose a vertical rectangular strip as the differential element of area (Fig. 1). Using a differential version of the parallel-axis theorem, you have
The element is symmetrical with respect to the x′ and y′ axes, so dI_{x′y′} = 0. From the geometry of the triangle, you can express the variables in terms of x and y.
\begin{aligned}y &=h\left(1-\frac{x}{b}\right) & d A &=y d x=h\left(1-\frac{x}{b}\right) d x \\\bar{x}_{e l} &=x & \bar{y}_{e l} &=\frac{1}{2} y=\frac{1}{2} h\left(1-\frac{x}{b}\right)\end{aligned}Integrating dI_{xy} from x = 0 to x = b gives you I_{xy}:
b. Product of Inertia \bar{I}_{x^{\prime \prime} y^{\prime \prime}}. The coordinates of the centroid of the triangle relative to the x and y axes are (Fig. 2 and Fig. 5.8A)
\bar{x} = \frac{1}{3}b \quad \quad \bar{y} = \frac{1}{3}hUsing the expression for I_{xy} obtained in part a, apply the parallel-axis theorem again:
REFLECT and THINK: An equally effective alternative strategy would be to use a horizontal strip element. Again, you would need to use the parallel-axis theorem to describe this strip, since each point in the strip would be a different distance from the y axis.
